Question

X²(4—x) / x²—10x+25<0Пжпжпжпж​

Answer 1

Ответ:

Объяснение:вот

Answer 2

To solve the inequality $\frac{x^2(4-x)}{x^2-10x+25} < 0$, you can follow these steps:

1. First, simplify the expression: $\frac{x^2(4-x)}{x^2-10x+25} = \frac{x^2(4-x)}{(x-5)^2}$

2. Now, determine the critical values by setting both the numerator and denominator equal to zero and solving for $x$:

   • Numerator: $x^2(4-x) = 0$ This gives you two critical values: $x = 0$ and $x = 4$.
   • Denominator: $(x-5)^2 = 0$ This gives you another critical value: $x = 5$.

3. Create a sign chart with these critical values and test intervals to determine where the expression is less than zero. We'll consider three intervals:

   • Interval 1: $(-∞, 0)$
   • Interval 2: $(0, 4)$
   • Interval 3: $(4, 5)$
   • Interval 4: $(5, +∞)$

4. Pick a test point within each interval and evaluate the expression:

   • Interval 1: Test $x = -1$ $\frac{x^2(4-x)}{(x-5)^2} = \frac{-1^2(4-(-1))}{(-1-5)^2} = \frac{5}{36}$
   • Interval 2: Test $x = 2$ $\frac{x^2(4-x)}{(x-5)^2} = \frac{2^2(4-2)}{(2-5)^2} = \frac{8}{9}$
   • Interval 3: Test $x = 4.5$ $\frac{x^2(4-x)}{(x-5)^2} = \frac{4.5^2(4-4.5)}{(4.5-5)^2} = -\frac{5.0625}{0.25} = -20.25$
   • Interval 4: Test $x = 6$ $\frac{x^2(4-x)}{(x-5)^2} = \frac{6^2(4-6)}{(6-5)^2} = \frac{36(-2)}{1} = -72$

5. Now, analyze the signs in each interval:

   • Interval 1: $\frac{5}{36} > 0$
   • Interval 2: $\frac{8}{9} > 0$
   • Interval 3: $-20.25 < 0$
   • Interval 4: $-72 < 0$

6. The expression is less than zero in intervals 3 and 4. Therefore, the solution to the inequality is: $-\infty < x < 4$ or $5 < x < \infty$

So, the values of $x$ for which $\frac{x^2(4-x)}{x^2-10x+25} < 0$ are $x < 4$ and $x > 5$.